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Chemistry: Post your doubts here!

Saad the Paki

Saad the Paki

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The answer is D. But the question asks what we can deduce from THESE observations. So how do we know that H2SO4 is a stronger oxidiser than iodine from these observations. And also, iodine is supposed to be a reducing agent and iodide is supposed to be an oxidising agent. It's the opposite here.Screenshot_2016-03-16-19-46-15.png
 
H

holoholo

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upload_2016-3-18_23-36-45.png

in 2014 may/june paper 42 mark scheme the optical isomers are given like this.

upload_2016-3-18_23-37-44.png

In the syllabus, it's given that potical isomers should be drawn like this.

My question is in the mark scheme why is the OH as a dashed line and then in the second isomer as a wedge but in the syllabus it is not ?
 

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The Sarcastic Retard

The Sarcastic Retard

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holoholo said:
View attachment 59711

in 2014 may/june paper 42 mark scheme the optical isomers are given like this.

View attachment 59713

In the syllabus, it's given that potical isomers should be drawn like this.

My question is in the mark scheme why is the OH as a dashed line and then in the second isomer as a wedge but in the syllabus it is not ?
Click to expand...
Maybe they have drawn two possiblities. If u see that other part is not optical isomer. So its two possiblities, either optical isomer of left or of right :)
 
qwertypoiu

qwertypoiu

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holoholo said:
View attachment 59711

in 2014 may/june paper 42 mark scheme the optical isomers are given like this.

View attachment 59713

In the syllabus, it's given that potical isomers should be drawn like this.

My question is in the mark scheme why is the OH as a dashed line and then in the second isomer as a wedge but in the syllabus it is not ?
Click to expand...
Both are fine. Key thing is to make sure the diagrams show the two molecules are non-superimposable.

Best to follow syllabus though.
 
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qwertypoiu

qwertypoiu

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holoholo said:
Are these complex ions ? and if so when we draw the structure how do we draw the dative bonding ?

View attachment 59722
Click to expand...
[PCl4]+ will be a tetrahedral structure, like CH4 or NH4+. I think showing the dative bonds is not important, but you could show one of the Cl atoms receiving a pair of electrons from P. (Just like N donates a lone pair to H in NH4+)

[PCl6]- will be an octahedral structure, like SF6. Again the coordination of the bonds shouldn't be important but you could show one Cl atom donating a pair of electrons to P.
 
H

holoholo

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qwertypoiu said:
[PCl4]+ will be a tetrahedral structure, like CH4 or NH4+. I think showing the dative bonds is not important, but you could show one of the Cl atoms receiving a pair of electrons from P. (Just like N donates a lone pair to H in NH4+)

[PCl6]- will be an octahedral structure, like SF6. Again the coordination of the bonds shouldn't be important but you could show one Cl atom donating a pair of electrons to P.
Click to expand...
How do we know if it is tetrahedral or square planar ?
 
qwertypoiu

qwertypoiu

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Saad the Paki said:
Anyone? The ans is A.View attachment 59700
Click to expand...
I'm guessing your trouble is with the right column.

  • You know there is a cyclohexene.
  • You know there are 20 carbon atoms.
  • You know there is an aldehyde group.
So you can deduce this:
upload_2016-3-19_19-15-31.png

The molecular formula for the above structure is C20H36O
We have 36-28 = 8 too many hydrogen atoms. By adding a double bond to the aliphatic chain on the right side, we can reduce the number of H atoms.
For every double bond added, TWO hydrogen atoms are removed. So we need to add 8/2 = 4 double bonds. It will look like this:
upload_2016-3-19_20-58-17.png

So that makes a total of 5 C=C double bonds.

P.S. the actual structure of 11-cis retinal is a bit different; it is a structural isomer of above but that doesn't really matter.
 
qwertypoiu

qwertypoiu

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holoholo said:
How do we know if it is tetrahedral or square planar ?
Click to expand...
Well this isn't a question about complexes firstly. So we should expect a tetrahedral structure.

Secondly, even if it were a question about complexes I'd suggest going with tetrahedral. Cisplatin is the only complex we need to know that is square planar, as far as I know.
 
qwertypoiu

qwertypoiu

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Saad the Paki said:
The answer is D. But the question asks what we can deduce from THESE observations. So how do we know that H2SO4 is a stronger oxidiser than iodine from these observations. And also, iodine is supposed to be a reducing agent and iodide is supposed to be an oxidising agent. It's the opposite here.View attachment 59691
Click to expand...
  • A strong oxidising agent is a specie that is really good at oxidising others.
  • So it is really good at getting reduced.
  • So it is really good at accepting electrons.
The strongest oxidising agent is the one that is the best at taking electrons and keeping it.
Let's look at the statements:

A - Phosphoric acid is a stronger oxidising agent than sulphuric acid?! No way! The experiments demonstrated that the SO4 2- ions were able to steal away the electrons from I- and keep it, but PO4 3- ions couldn't do that. So this statement is opposite. Sulphuric acid should be stronger oxidising agent.

B - Phosphoric acid better oxidising agent that iodine? Wait a sec. Where does I2 even come from? We can imagine this equilibrium existing:
I2 + 2e <----> I-​
I2 is an oxidising agent because it can take away electrons from others to become I-. One way to measure its oxidizing ability would be to see if it is able to keep these electrons and remain in the form of I-. If a specie comes along and oxidises (snatches away electrons from) I-, it means I2 as an oxidising agent was not strong enough to keep these electrons and remain I-. Hope this makes sense.

So was phosphoric acid strong enough to take away electrons from I- to form I2? Nope. So I2 is actually a stronger oxidising agent than H3PO4.

C - For the same reason as above, since H2SO4 was not able to oxidise Cl- to Cl2, Cl2 is the stronger oxidising agent.

D - In this case, sulphuric acid WAS able to oxidise I- to I2. So H2SO4 is the stronger oxidising agent (when compared to I2)
 
qwertypoiu

qwertypoiu

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Saad the Paki said:
View attachment 59687
I think both A and D are possible answerstandards but it's D in the ms. :/
Click to expand...
Hydrogen bonds between molecules is an intermolecular force of attraction. Intermolecular forces are best shown by looking at energy required to vaporize, since the process makes the molecules as far away from each other as possible (thus destroying all intermolecular forces)

Bond dissociation energy (which is same thing as bond energy I assume) will just show energy required to break H-X bond (where X is halogen like Cl)
These values are completely irrelevant, since these H-X bonds aren't even being broken. I hope you understand this. In a reaction, H-X bonds could be broken, but during vaporization, HX will remain in its molecular state. Only the molecules will move further apart from one another.
 
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